Shadow Posted December 10, 2011 Share Posted December 10, 2011 Given the matrix [math] A = \left( \begin{array}{cccc} 2 & 1 & 0 & -1\\ 0 & 2 & -1 & -1\\ 0 & 0 & 3 & 1\\ 0 & 0 & -1 & 1 \end{array}\right) [/math], find its Jordan matrix and the matrix T, for which [math]A=TJT^{-1}[/math]. All four of the eigenvalues are 2, and the Jordan matrix will consist of two Jordan blocks; [math] J = \left( \begin{array}{cccc} 2 & 1 & 0 & 0\\ 0 & 2 & 0 & 0\\ 0 & 0 & 2 & 1\\ 0 & 0 & 0 & 2 \end{array}\right) [/math]. Thus, if we label the eigenvectors [math]h_1[/math], [math]h_2[/math], [math]h_3[/math], [math]h_4[/math], we have [A].[[math]h_1[/math] | [math]h_2[/math] | [math]h_3[/math] | [math]h_4[/math] ] = [[math]h_1[/math] | [math]h_2[/math] | [math]h_3[/math] | [math]h_4[/math] ].[J], and from this A[math]h_1[/math] = 2[math]h_1[/math] A[math]h_2[/math] = [math]h_1[/math] + 2[math]h_2[/math] A[math]h_3[/math] = 2[math]h_3[/math] A[math]h_4[/math] = [math]h_3[/math] + 2[math]h_4[/math] But this would mean that [math]h_1[/math] = [math]h_3[/math] and thus [math]h_2[/math] = [math]h_4[/math], which is nonsense. And I'm stumped as to what to do about it. Any ideas? Link to comment Share on other sites More sharing options...
Schrödinger's hat Posted December 10, 2011 Share Posted December 10, 2011 Degenerate nxn matrices don't always have n linearly independant eigenvectors. Link to comment Share on other sites More sharing options...
Shadow Posted December 11, 2011 Author Share Posted December 11, 2011 (edited) I did not know that, but nevertheless this one does; the solution is [math] T = \left( \begin{array}{cccc} -1 & 0 & 0 & 0\\ -1 & 0 & -1 & 0\\ 1 & 0 & 1 & 1\\ -1 & 1 & -1 & 0 \end{array}\right) [/math] Edited December 11, 2011 by Shadow Link to comment Share on other sites More sharing options...
DrRocket Posted December 12, 2011 Share Posted December 12, 2011 Degenerate nxn matrices don't always have n linearly independant eigenvectors. If you had n linearly independent eigenvectors then the matrix would be diagonalizable and you would not need Jordan normal form. Link to comment Share on other sites More sharing options...
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